Question

What is \( \frac{\cos \theta}{1 - \tan \theta} + \frac{\sin \theta}{1 - \cot \theta} \) equal to?

• A. \( \sin \theta - \cos \theta \)
• B. \( 2 \sin \theta \)
• C. \( \sin \theta + \cos \theta \)
• D. \( 2 \cos \theta \)

Hint:

Use basic trigonometric identities such as \( \tan \theta = \frac{\sin \theta}{\cos \theta} \) and \( \cot \theta = \frac{\cos \theta}{\sin \theta} \) to simplify complex fractions. Also, factor identities like \( \cos^2 \theta - \sin^2 \theta = (\cos \theta + \sin \theta)(\cos \theta - \sin \theta) \) are very helpful.

Answer 1

The Correct Option is C

Step 1: Simplify the first term
\[ \frac{\cos \theta}{1 - \tan \theta} = \frac{\cos \theta}{1 - \frac{\sin \theta}{\cos \theta}} = \frac{\cos \theta}{\frac{\cos \theta - \sin \theta}{\cos \theta}} = \frac{\cos^2 \theta}{\cos \theta - \sin \theta} \] Step 2: Simplify the second term
\[ \frac{\sin \theta}{1 - \cot \theta} = \frac{\sin \theta}{1 - \frac{\cos \theta}{\sin \theta}} = \frac{\sin \theta}{\frac{\sin \theta - \cos \theta}{\sin \theta}} = \frac{\sin^2 \theta}{\sin \theta - \cos \theta} = - \frac{\sin^2 \theta}{\cos \theta - \sin \theta} \] Step 3: Add the two simplified expressions
\[ \frac{\cos^2 \theta}{\cos \theta - \sin \theta} - \frac{\sin^2 \theta}{\cos \theta - \sin \theta} = \frac{\cos^2 \theta - \sin^2 \theta}{\cos \theta - \sin \theta} \] Step 4: Use identity
\[ \cos^2 \theta - \sin^2 \theta = (\cos \theta + \sin \theta)(\cos \theta - \sin \theta) \] Step 5: Final simplification
\[ \frac{(\cos \theta + \sin \theta)(\cos \theta - \sin \theta)}{\cos \theta - \sin \theta} = \cos \theta + \sin \theta \]